HackerEarth Round Table Meeting problem solution

In this HackerEarth Round Table Meeting problem solution, There are N students in a round table meeting. Each student belongs to a university given in the array A[i], which denotes the university that the ith student belongs to.

There are Q queries of the form x y, denoting two universities. The answer to each query is the minimum time taken by any one of the student from these universities to meet each other.

#include<bits/stdc++.h>
using namespace std;
#define ll long long 
#define modu 1000000007
#define pb push_back
vector<int> v[1000001];
int bs(int numElems,int b,int target)
{
    int low = 0, high = numElems; 
    while (low != high) 
    {
        int mid = (low + high) / 2;
        if (v[b][mid] <= target) 
        {
            low = mid + 1;
        }
        else 
        {
            high = mid;
        }
    }
    return high;
}
int mod(int a)
{
    if(a<0)
    return -a;
    else
    return a;
}
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int n,q;
    cin>>n>>q;
    int A[200005];
    for(int i=0;i<n;i++)
    {
        cin>>A[i];
    }
    for(int i=n;i<2*n;i++)
    {
        A[i]=A[i-n];
    }
    for(int i=0;i<2*n;i++)
    {
        v[A[i]].pb(i);
    }
    while(q--)
    {
        int mindiff=200000;
        int a,b;
        cin>>a>>b;
        for(int i=0;i<v[a].size();i++)
        {
            int y=bs(v[b].size(),b,v[a][i]);
            int val=(v[b][y]-v[a][i]);
            mindiff=min(mindiff,mod(val));
            if(y!=0)
            mindiff=min(mindiff,mod(v[a][i]-v[b][y-1]));
        }
        cout<<mindiff/2<<endl;
    }
}